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To solve the equation [tex]\(\tan^2(x) + 2\sqrt{3}\tan(x) + 3 = 0\)[/tex] for [tex]\(x\)[/tex] in the range [tex]\(\pi \leq x \leq 2\pi\)[/tex], we will follow these steps:
1. Solve the quadratic equation for [tex]\(\tan(x)\)[/tex]:
[tex]\[ t = \tan(x) \][/tex]
This simplifies the given equation to:
[tex]\[ t^2 + 2\sqrt{3}t + 3 = 0 \][/tex]
Here, we can solve for [tex]\(t\)[/tex] using the quadratic formula, [tex]\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 2\sqrt{3}\)[/tex], and [tex]\(c = 3\)[/tex].
Substituting these values into the quadratic formula:
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{12 - 12}}{2} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{0}}{2} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3}}{2} \][/tex]
[tex]\[ t = -\sqrt{3} \][/tex]
So, the solution for [tex]\(\tan(x)\)[/tex] is:
[tex]\[ \tan(x) = -\sqrt{3} \][/tex]
2. Find the corresponding [tex]\(x\)[/tex] values in the range [tex]\(\pi \leq x \leq 2\pi\)[/tex]:
We need to find [tex]\(x\)[/tex] such that:
[tex]\[ \tan(x) = -\sqrt{3} \][/tex]
To find these values, we look for the arctangent (or inverse tangent) solutions and consider the periodicity of the tangent function. The basic solution for [tex]\(\tan(x) = -\sqrt{3}\)[/tex] within one period can be found using:
[tex]\[ x = \arctan(-\sqrt{3}) + k\pi \][/tex]
where [tex]\( k \)[/tex] is an integer. Specifically,
[tex]\[ \arctan(-\sqrt{3}) \approx -\frac{\pi}{3} \][/tex]
3. Adjust for the given range [tex]\(\pi \leq x \leq 2\pi\)[/tex]:
For the solutions to fall within [tex]\(\pi \leq x \leq 2\pi\)[/tex], we consider adding [tex]\( k\pi \)[/tex] to our arctan solution from above.
- For [tex]\(k = 1\)[/tex]:
[tex]\[ x = -\frac{\pi}{3} + \pi = \frac{2\pi}{3} \][/tex]
This value does not fall within the specified range.
- For [tex]\(k = 2\)[/tex]:
[tex]\[ x = -\frac{\pi}{3} + 2\pi = \frac{5\pi}{3} \][/tex]
This value does fall within the range [tex]\(\pi \leq x \leq 2\pi\)[/tex].
Thus, the suitable [tex]\(x\)[/tex] value falling in the specified range is:
[tex]\[ x = \frac{5\pi}{3} \approx 5.236 \][/tex]
So, the complete solution indicates that the value of [tex]\(x\)[/tex] in the interval [tex]\(\pi \leq x \leq 2\pi\)[/tex] that satisfies the equation [tex]\(\tan^2(x) + 2\sqrt{3}\tan(x) + 3 = 0\)[/tex] is [tex]\( x = \frac{5\pi}{3} \approx 5.236 \)[/tex].
1. Solve the quadratic equation for [tex]\(\tan(x)\)[/tex]:
[tex]\[ t = \tan(x) \][/tex]
This simplifies the given equation to:
[tex]\[ t^2 + 2\sqrt{3}t + 3 = 0 \][/tex]
Here, we can solve for [tex]\(t\)[/tex] using the quadratic formula, [tex]\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 2\sqrt{3}\)[/tex], and [tex]\(c = 3\)[/tex].
Substituting these values into the quadratic formula:
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{12 - 12}}{2} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{0}}{2} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3}}{2} \][/tex]
[tex]\[ t = -\sqrt{3} \][/tex]
So, the solution for [tex]\(\tan(x)\)[/tex] is:
[tex]\[ \tan(x) = -\sqrt{3} \][/tex]
2. Find the corresponding [tex]\(x\)[/tex] values in the range [tex]\(\pi \leq x \leq 2\pi\)[/tex]:
We need to find [tex]\(x\)[/tex] such that:
[tex]\[ \tan(x) = -\sqrt{3} \][/tex]
To find these values, we look for the arctangent (or inverse tangent) solutions and consider the periodicity of the tangent function. The basic solution for [tex]\(\tan(x) = -\sqrt{3}\)[/tex] within one period can be found using:
[tex]\[ x = \arctan(-\sqrt{3}) + k\pi \][/tex]
where [tex]\( k \)[/tex] is an integer. Specifically,
[tex]\[ \arctan(-\sqrt{3}) \approx -\frac{\pi}{3} \][/tex]
3. Adjust for the given range [tex]\(\pi \leq x \leq 2\pi\)[/tex]:
For the solutions to fall within [tex]\(\pi \leq x \leq 2\pi\)[/tex], we consider adding [tex]\( k\pi \)[/tex] to our arctan solution from above.
- For [tex]\(k = 1\)[/tex]:
[tex]\[ x = -\frac{\pi}{3} + \pi = \frac{2\pi}{3} \][/tex]
This value does not fall within the specified range.
- For [tex]\(k = 2\)[/tex]:
[tex]\[ x = -\frac{\pi}{3} + 2\pi = \frac{5\pi}{3} \][/tex]
This value does fall within the range [tex]\(\pi \leq x \leq 2\pi\)[/tex].
Thus, the suitable [tex]\(x\)[/tex] value falling in the specified range is:
[tex]\[ x = \frac{5\pi}{3} \approx 5.236 \][/tex]
So, the complete solution indicates that the value of [tex]\(x\)[/tex] in the interval [tex]\(\pi \leq x \leq 2\pi\)[/tex] that satisfies the equation [tex]\(\tan^2(x) + 2\sqrt{3}\tan(x) + 3 = 0\)[/tex] is [tex]\( x = \frac{5\pi}{3} \approx 5.236 \)[/tex].
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